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How to find the velocity of a particle which has mass $m$ and energy $E$ considering the non-relativistic and the extreme relativistic limits?

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marked as duplicate by Daniel Griscom, ACuriousMind, John Rennie, user36790, Community Jan 26 '16 at 10:34

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Non-relativistic: The kinetic energy of a particle is given by $\frac {mv^2} 2$ . Assuming no rotation, that's it for kinetic energy. $E$ is kinetic energy plus potential energy of an object. There is also its energy due to its mass $m_0c^2$, but for a non-relativistic particle this is so large we can neglect it. So if $E$ is all kinetic, $v = \sqrt(\frac{2E} m)$ and you will have $v$ within a plus or minus sign.

Relativistic: $E$ is given by $m_0\gamma c^2$, where $\gamma = \frac {1} {\sqrt(1-\frac {v^2} {c^2})}$. The relativistic kinetic energy is $(\gamma -1)m_0c^2$ ($m_0$ is the object's rest mass). So if you know $E$ and $m_0$, you can calculate $\gamma$ using $E = m_0\gamma c^2$ and find $v$ using the equation for $\gamma$.

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  • $\begingroup$ The relativistic expression for the total energy is $E^2 = p^2c^2 + m^2c^4$ where $p = \gamma mv$. The kinetic energy is not a very useful concept in relativity, but if you insisted on using it it would be $KE = mc^2\left(\sqrt{\frac{\gamma^2v^2}{c^2}+1} - 1\right)$. $\endgroup$ – John Rennie Jan 26 '16 at 6:40

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